Both R1 and R2 are equivalence relations
Neither R1 nor R2 is an equivalence relation
R1 is an equivalence relation but R2 is not
R2 is an equivalence relation but R1 is not
The correct answer is (B) : Neither R1 nor R2 is an equivalence relation
R1 = {(a, b) ∈ N × N : |a – b| ≤ 13} and
R2 = {(a, b) ∈ N × N : |a – b| ≠ 13}
In R1: ∵ |2 – 11| = 9 ≤ 13
∴ (2, 11) ∈ R1 and (11, 19) ∈ R1 but
(2, 19) ∉ R1
∴ R1 is not transitive
Hence R1 is not equivalence
In R2 : (13, 3) ∈ R2 and (3, 26) ∈ R2 but
(13, 26) ∉ R2 (∵ |13 – 26| = 13)
∴ R2 is not transitive
Hence R2 is not equivalence.
In mathematics, a set is a well-defined collection of objects. Sets are named and demonstrated using capital letter. In the set theory, the elements that a set comprises can be any sort of thing: people, numbers, letters of the alphabet, shapes, variables, etc.
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The items existing in a set are commonly known to be either elements or members of a set. The elements of a set are bounded in curly brackets separated by commas.
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The cardinal number, cardinality, or order of a set indicates the total number of elements in the set.
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